Let y = (x-3)/(x+7) + 1/3
Then this equation can be expressed in the form (x + a)(y + b) = c for some constants a, b, and c . Enter your answer in the form " a, b, c".
+2666 First, note that \((x+a)(y+b) = xy + xb + ay + ab\)
Start by simplifying the RHS:
\(={x - 3 \over x + 7} + {1 \over 3}\)
\(= { 3(x - 3) \over 3(x + 7) } + {x + 7 \over 3x + 21}\)
\(= { 3x-9 \over 3x+21 } + {x + 7 \over 3x + 21}\)
\(= { 4x - 2\over 3x + 21}\)
Now, we have the equation \(y = {4x - 2\over 3x + 21}\)
Then, cross multiply to get \(3xy + 21y = 4x - 2\)
Subtract 4x from both sides: \(3xy + 21y - 4x =- 2\)
Divide by 3: \(xy + 7y - {4 \over 3}x = -{2 \over 3}\)
Now, recall the factored form. We know that \(ay = 7y\) and \(bx = -{4 \over 3}x\), meaning \((a,b) = \ (7, -{4 \over 3})\)
But, remember that we need to add \(a \times b = 7 \times -{4 \over 3} = -{28 \over 3}\), so we can factor. Thus, \(c = -{2 \over 3} -{28 \over 3} = -10\)
This means it is \(\color{brown}\boxed{(x + 7) (y - {4 \over 3})=-10}\)
